Question

A 60-MVA, 69.3-kV, three-phase synchronous generator has a synchronous reactance of 15 Ω/phase and negligible armature resistance.
-a. The generator is delivering rated power at 0.8 power factor lagging at the rated terminal voltage to an infinite bus bar. Determine the magnitude of the generated emfper phase and the power angle δ.
-b. If the generated emfis 36 kV per phase, what is the maximum three-phase power that the generator can deliver before losing its synchronism?
Winter

Answer 1

bA z - Phi synchronous generator , 60MVA V_L = 69.3 kv and = 15Ohm/ph V_ph = V_L/squareroot 3 = 69.3 times 10^3/squareroot 3 = 39837.17v I_L = I_ph = 60 times 10^6/squareroot 3 V_L = 60 times 10^6/Sqaureroot 3 times 69.3 times 10^3 I_L = I_ph = 499.87 A a) Generated emf per phase at 0.8 pf lagging load E_ph = V_ph angle0 + I_a angle Phi Z_x angle theta E_ph = 39837.17 + 499.87 angle -36.87 degree times 15 angle 90 degree = 39837.17 + 7498.05 angle 53.13 degree v power angle delta = 7.705 degree b) p = 3EV/x_s sin delta P_max = 3EV/x_s E = 36 kv/ph P_max = 3 times 36 times 10^3 times 39.837 times 10^3/15 P_max = 286.826 mw